The geometric idea of the dot product has been touched upon, but there is a vast generalization of this product in geometric algebra, the algebra of not only oriented lines (vectors) but planes, volumes, and more (called blades).
In geometric algebra, we have a generalized dot product of a vector $a$ and another blade $B$ denoted $a \cdot B$. This product has a simple geometric interpretation as the part of $B$ orthogonal to the projection of $a$, with magnitude $|a||B|| \cos \theta|$, where $\theta$ is the angle $a$ forms with its projection in $B$.
(Note: as a point of fact, $a \cdot B$ is orthogonal to $a$ also, not just its projection into $B$, but thinking about this leads to some difficulties, while thinking about what's perpendicular to the projection does not.)
If $B$ is a 2-blade (also called a bivector), then you should be able to imagine this directly: if $a$ lies entirely in $B$, then $a \cdot B$ is just the vector perpendicular to $a$ in $B$. If $a$ does not lie entirely in $B$, then it can be decomposed into a tangential part and a normal part. We throw away the normal part, and the previous logic applies for the tangential part.
If $B$ is a 3-blade (a trivector), then in 3d space $a$ must lie in $B$ (for there is no 3d volume that a vector does not help span), and the product $a \cdot B$ is the "Hodge dual", or the plane perpendicular to $a$.
In this light, the dot product of vectors may actually be the most non-intuitive part of this reasoning. When you take the dot product, there's only a scalar left--there's no vector or other higher dimensional object left to be orthogonal to $a$. Again, this is why I emphasize that $a \cdot B$ is the part of $B$ orthogonal to the projection of $a$ onto $B$. When $B$ is a vector, it's clear there is no other vector or anything else that can be orthogonal to the projection of $a$, for $B$ and that projection are parallel, so the result is necessarily just a scalar.